Question

rezolvati in r ecuatia |-6(x+3)|>4

Answer 1

Din proprietatile modulului avem:
$|-6(x+3)|= |-6|\cdot|x+3|=6\cdot|x+3|$
Vom imparti modulul in doua cazuri:
I)
$x+3\geq0 \Leftrightarrow x\geq-3 \Leftrightarrow x\in[-3,\infty )\\ 6\cdot|x+3|\ \textgreater \ 4 \Leftrightarrow 6(x+3)\ \textgreater \ 4 \Leftrightarrow \\ 6x+18\ \textgreater \ 4 \Leftrightarrow 6x\ \textgreater \ -14 \Leftrightarrow x\ \textgreater \ \frac{-14}{6} \Leftrightarrow x\in(\frac{-14}{6},\infty)\\ x\in([-3,\infty)\cap(\frac{-14}{6},\infty)) \Leftrightarrow \underline{x\in(-\frac{14}{6},\infty)}$
II)
[tex]x+3\ \textless \ 0 \Leftrightarrow x\ \textless \ -3 \Leftrightarrow x\in(-\infty,-3)\\ 6\cdot|x+3|\ \textgreater \ 4 \Leftrightarrow 6(-x-3)\ \textgreater \ 4 \\ \Leftrightarrow -6x-18\ \textgreater \ 4 \Leftrightarrow -6x\ \textgreater \ 22\\ \Leftrightarrow x\ \textless \ \frac{-22}{8} \Leftrightarrow x\in(-\infty, \frac{-22}{8}) \\ \\ x\in((-\infty,-3)\cap(-\infty, \frac{-22}{8})) \Leftrightarrow \underline{x\in(-\infty,-3)}[/tex]
De aici, avem ca:
$x\in((-\frac{14}{6},\infty)\cup(-\infty,-3)) \Leftrightarrow\\ x\in \mathbb{R}\setminus [-3,\frac{14}{6}]$